Error in Argument with Reduced Mayer-Vietoris for Disjoint Union I was trying to prove something and came across an apparent contradiction. I am sure there is a very basic mistake somewhere in my line of reasoning, but I can't find it and would appreciate it if someone else can.
So let $X$ and $Y$ be contractible spaces and $Z=X\sqcup Y$ their disjoint union. Now consider the reduced version of the Mayer-Vietoris sequence. We have:
$$
\cdots\to\tilde{H}_0(X\cap Y)\to \tilde{H}_0(X)\oplus\tilde{H}_0(Y)\to\tilde{H}_0(Z)\to 0
$$ 
Now $X\cap Y$ is empty, and $\tilde{H}_0(X)=\tilde{H}_0(Y)=0$ because we assume $X$ and $Y$ to be contractible. But $Z$ has two components, namely $X$ and $Y$, so $\tilde{H}_0(Z)=\mathbb{Z}$.
Thus, the Mayer-Vietoris sequence is given by
$$
\cdots\to 0\to 0\to\mathbb{Z}\to 0
$$
which cannot possibly be exact. So where am I going wrong?
 A: The problem is you pass to reduced homology before forming $H_0X\oplus H_0Y$. I'll reference Hatcher's statement and proof (pg 149) in the context of singular homology, since it is particularly lucid. Observe that the existence of the Mayer-Vietor sequence is established by Hatcher by forming the short exact sequence of chain complexes
$$0\rightarrow C_*(X\cap Y)\rightarrow C_*(X)\oplus C_*(Y)\rightarrow C_*(X\cup Y)\rightarrow 0\qquad (\ast)$$
and computing the resulting long-exact sequence of homology groups. The result is the long exact sequence
$$\dots\rightarrow H_n(X\cap Y)\rightarrow H_n(X)\oplus H_n(Y)\rightarrow H_n(X\cup Y)\rightarrow H_{n-1}(X\cap Y)\rightarrow\dots$$
in unreduced homology, which ends 
$$\dots\rightarrow H_0(X\cap Y)\rightarrow H_0(X)\oplus H_0(Y)\rightarrow H_0(X\cup Y)\rightarrow 0$$
If you prefer to work with reduced homology then you have to use the reduced homology of the sequence labelled $(\ast)$. i.e. you have to treat $C_*(X)\oplus C_*(Y)$ as one object.
Recall that in the singular context, the reduced homology $\widetilde H_n(K)$ is defined (pg 110) to be the $n^{th}$ homology group of the augemented chain complex
$$\dots\rightarrow C_n(K)\rightarrow C_{n-1}(K)\rightarrow\dots\rightarrow C_1(K)\rightarrow C_0(K)\xrightarrow\epsilon \mathbb{Z}\rightarrow 0$$
Thus the correct way to take the reduced homology of the chain complex $C_\ast(X)\oplus C_*(Y)$ is to take the homology of the chain complex ending
$$\dots\rightarrow C_1(X)\oplus C_1(Y)\rightarrow C_0(X)\oplus C_0(Y)\xrightarrow{\epsilon\oplus\epsilon}\mathbb{Z}\oplus\mathbb{Z}\rightarrow0$$
Now the chain complex $(\ast)$ ends in degree $-1$ as
$$0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\oplus\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow0.$$
For example, in the example you create you find exactness of 
$$0\rightarrow \widetilde H_0(X)\oplus\widetilde H_0(Y)\rightarrow\widetilde H_0(X\sqcup Y)\rightarrow \mathbb{Z}\rightarrow 0$$
where of course $\widetilde H_0(X)\oplus\widetilde H_0(Y)=0$.
This is discussed by hatcher on pg 150. You can check that you get the correct answer when using this correct definition.
A: Here is an alternate approach. I find that exercise 38 on p. 159 of Hatcher is a good way to derive the Mayer-Vietoris sequence: there is a commutative diagram
$$
\begin{array}{ccccccccc}
\to & H_{n+1}(X, X \cap Y) & \to & H_n (X \cap Y) & \to & H_n(X) & \to & H_n(X, X \cap Y) & \to \\
 & \downarrow^\cong & & \downarrow & & \downarrow & & \downarrow^\cong \\
\to & H_{n+1}(X \cup Y, Y) & \to & H_n (Y) & \to & H_n(X \cup Y) & \to & H_n(X \cup Y, Y) & \to
\end{array}
$$
Because of the isomorphisms at the end, a diagram chase (this is Hatcher's exercise 38) produces the Mayer-Vietoris sequence.
Since you also have a commutative diagram with reduced homology, you also get a Mayer-Vietoris sequence there, but as always with reduced homology, you have to be careful with empty sets: the reduced homology of the empty set is $H_q(\emptyset) = 0$ if $q \geq 0$ but $H_{-1}(\emptyset) \cong \mathbb{Z}$ — this is also the same as the augmented chain complex for the empty set. The tail end of the Mayer-Vietoris sequence is
$$
\to \widetilde{H}_0(X \cap Y) \to \widetilde{H}_0(X) \oplus \widetilde{H}_0(Y) \to \widetilde{H}_0(X \cup Y) \to \widetilde{H}_{-1}(X \cap Y) \to
$$
so the $(-1)$-dimensional homology of the empty set can enter into play.
